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Civil War physics homework

  1. Feb 24, 2006 #1
    As a hobby, you like to participate in reenactments of Civil War battles. Civil War cannons were muzzle loeaded, meaning that the gunpowder and the cannonball were inserted into the output end of the muzzle, then tamped into place with a long plunger. To recreate the authenticity of muzzle-loaded cannons, but without the danger of real cannons, Civil War buffs have invented a spring-powered cannon that fires a 1.0kg plastic ball. A spring, with constant 3000N/m, is mounted at the back of the barrel. You place a ball in the barrel, then use a long plunger to press the ball against the spring and lock the spring into place, ready for firing. In order for the latch to catch, the ball has to be moving at a speed of at least 2.0m/s when the spring has been compressed 30cm. The coefficient of friction of the ball in the barrel is 0.30. The plunger doens't touch the sides of the barrel.

    a) if you push the plunger with a constant force, what is the minimum force that you must use to compress and latch the spring? You can assume that no effort was required to push the ball down the barrel to where it first contacts the spring.

    b) what is the cannon's muzzle velocity if the ball travels a total distance of 1.5m to the end of the barrel?
     
  2. jcsd
  3. Feb 24, 2006 #2
    For a did this this
    Fy=0=Fn-mg
    Ffr=(mu)mg=2.94N
    delta(K)=Wc+Wdiss+Wext
    2J=-135J-0.882J+Wext
    Wext=137.87J
    W=Fd
    F=459N
     
  4. Feb 24, 2006 #3
    for b i'm a little bit confused I believe that:
    delta(K)=Delta(U)+Wdiss
    so velocity as soon as it leaves the spring is
    vf=16.5m/s
     
  5. Feb 24, 2006 #4

    Hootenanny

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    I believe you need the angle of elevation of the cannon.
     
  6. Feb 24, 2006 #5
    hmm... i think you just assume it is parallel to the ground
     
  7. Feb 24, 2006 #6

    Hootenanny

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    What is U and K in your equation?
     
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